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Exer. 3546: Sketch the graph of f | Ch 5.5 - 40

Algebra and Trigonometry with Analytic Geometry | 12th Edition | ISBN: 9780495559719 | Authors: Earl Swokowski, Jeffery A. Cole ISBN: 9780495559719 225

Solution for problem 40 Chapter 5.5

Algebra and Trigonometry with Analytic Geometry | 12th Edition

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Algebra and Trigonometry with Analytic Geometry | 12th Edition | ISBN: 9780495559719 | Authors: Earl Swokowski, Jeffery A. Cole

Algebra and Trigonometry with Analytic Geometry | 12th Edition

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Problem 40

Exer. 3546: Sketch the graph of f

Step-by-Step Solution:
Step 1 of 3

Calculus I Chapters 1.1 and 1.2 Chapter 1.1 – An Introduction to Calculus  Formulas to remember 2 2 o ( − + = − o ( + )2 = + 2 + 2 2 2 2 o ( − ) = − 2 − −± −4 o = 2 1.1 Preview of Calculus  Calculus is the mathematics of change Pre-Cal Cal Δ Slope = Δ Area Of rectangle = ℎ 2) the Tangent Line Problem Finding a tangent line to a curve @ point P is equivalent to finding the slope of the tangent line at P. - We will use the secant line: a line through the point P of the tangent and a point Q also on the curve. ( ) 2 For = = 1 = 1 2 1.1 = 1.1 = 1.21 1(1.1, 1.1 ) = 1.1,1.21 ) (1.5 , 1.5 ) = (1.5,2.25) 2 3(1.01, 1.01 ) = 1.01,1.0201 ) (1.001, 1.001 ) = 1.001,1.002 ) 4 Slope of P, Q + ∆ − ) + ∆ − ) = = ( + ∆ − ∆ = 1 ∆ = 1 − 1.1 = 0.1 1 + 0.1 − 1) 1 = 0.1 1.1 − 1) 1.21 − 1 0.21 = = = = 2.1 1 0.1 0.1 0.1 1.01 − 1 ) 3 = 0.01 1.0201 − 1 0.0201 3 = 0.01 = 0.01 = 2.01 1.001 − 1 ) 4 = 0.001 1.002 − 1 0.002 4 = = = 2 0.001 0.001 2 = − 1 : = 1 − 1 0 = 1, 1 = , −1 = 1 = , = 2 + 1 ( ) = 2 + 1 )2= 4 + 4 + 4 ( ) = 2 + 1 Chapter 1.2 Finding Limits Graphically and Numerically 1) The informal definition of a limit a. b. ℎ c. If () becomes extremely close to the number and approaches from either side, then we say: ℎ ,ℎ . And we write graphically → = Ex: = −2 = −2 = 1 = 2 −4 (+2)(−2) +2 1 lim = →2 4 lim = →−2 Numerically: 1.9 1.99 1.999 2 2.001 2.01 2.1 () 0.256 0.251 0.250 0.25 0.2499 0.2494 0.2439 2) Learning and using the formal definition of a limit → = This means that for each > 0, there exists a > 0 If 0 < − < , then | − | < and are positive numbers 0 < − < either is in the interval ( − , + ) or is in ( − , or ( + ,) Ex: using the term and definition = + 2 = 4 Lim + 2 = 6 →4 Find that satisfies: 0 < − < implies − < ( ) | | − 6 = + 2 − = | − 4|| If we want − 4 < 3, must choose ) 3) Learn different ways that a limit can fail to exist I) Behavior that differs from the left and the right Y-Values 2.5 2 1.5 1 0.5 0 -3 -2 -1 0 1 2 3 4 5 -0.5 -1 -1.5 -2 -2.5 II) Unbounded behavior III) Oscillating behavior = sin( )1 1 lim sin( ) = →0

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Chapter 5.5, Problem 40 is Solved
Step 3 of 3

Textbook: Algebra and Trigonometry with Analytic Geometry
Edition: 12
Author: Earl Swokowski, Jeffery A. Cole
ISBN: 9780495559719

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Exer. 3546: Sketch the graph of f | Ch 5.5 - 40